If `A={x|cosxgt-(1)/(2)and0lexlepi}` and `B={x|sinxgt(1)/(2)and(pi)/(3)lexlepi}`and if `"pilambdaleAnnBltpimu`, the value of `(lambda + mu)` is

Explanation

To solve the problem step by step, we will analyze the sets \( A \) and \( B \) defined by the given inequalities and then find the intersection of these sets. Finally, we will determine the values of \( \lambda \) and \( \mu \) based on the intersection and compute \( \lambda + \mu \).

### Step 1: Define Set \( A \)
Set \( A \) is defined as:
\[ A = \{ x \mid \cos x > -\frac{1}{2} \text{ and } 0 \leq x \leq \pi \} \]
To find the values of \( x \) for which \( \cos x > -\frac{1}{2} \), we first find where \( \cos x = -\frac{1}{2} \).

The angles where \( \cos x = -\frac{1}{2} \) in the interval \( [0, \pi] \) are:
\[ x = \frac{2\pi}{3} \]
Thus, \( \cos x > -\frac{1}{2} \) for:
\[ 0 \leq x < \frac{2\pi}{3} \]
So, the set \( A \) can be expressed as:
\[ A = [0, \frac{2\pi}{3}) \]

### Step 2: Define Set \( B \)
Set \( B \) is defined as:
\[ B = \{ x \mid \sin x > \frac{1}{2} \text{ and } \frac{\pi}{3} \leq x \leq \pi \} \]
To find the values of \( x \) for which \( \sin x > \frac{1}{2} \), we first find where \( \sin x = \frac{1}{2} \).

The angles where \( \sin x = \frac{1}{2} \) in the interval \( [0, \pi] \) are:
\[ x = \frac{\pi}{6} \]
Thus, \( \sin x > \frac{1}{2} \) for:
\[ \frac{\pi}{6} < x < \frac{5\pi}{6} \]
However, we also have the restriction \( \frac{\pi}{3} \leq x \leq \pi \). Therefore, the set \( B \) can be expressed as:
\[ B = \left[\frac{\pi}{3}, \frac{5\pi}{6}\right) \]

### Step 3: Find the Intersection \( A \cap B \)
Now we find the intersection of sets \( A \) and \( B \):
\[ A \cap B = [0, \frac{2\pi}{3}) \cap \left[\frac{\pi}{3}, \frac{5\pi}{6}\right) \]
The intersection occurs in the interval:
\[ A \cap B = \left[\frac{\pi}{3}, \frac{2\pi}{3}\right) \]

### Step 4: Determine \( \lambda \) and \( \mu \)
We are given that:
\[ \pi \lambda \leq A \cap B < \pi \mu \]
From the intersection \( \left[\frac{\pi}{3}, \frac{2\pi}{3}\right) \), we can deduce:
- The lower bound \( \frac{\pi}{3} \) corresponds to \( \pi \lambda \), hence:
\[ \lambda = \frac{1}{3} \]
- The upper bound \( \frac{2\pi}{3} \) corresponds to \( \pi \mu \), hence:
\[ \mu = \frac{2}{3} \]

### Step 5: Calculate \( \lambda + \mu \)
Now, we can compute:
\[ \lambda + \mu = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1 \]

### Final Answer
Thus, the value of \( \lambda + \mu \) is:
\[ \boxed{1} \]